In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is

where D=5.00 m is the distance of the screen from the slits, and

is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:

And from the relationship between frequency and wavelength,

, we can find the frequency of the light:
C. element only one substance
Answer:
D. only briefly while being connected or disconnected.
Explanation:
As we know that transformer works on the principle of mutual inductance
here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil
So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.
Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always
So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output
so correct answer will be
D. only briefly while being connected or disconnected.
Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :


E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.